Question

what is the formula of the compound between calcium and sulfur that has the percent composition 55.6% (by mass) calcium and 44.4% (by mass) sulfur?
a cas
b cas₄
c ca₂s
d cas₂

Explanation:

Step1: Calculate moles of Ca

Assume 100g of the compound. Mass of Ca = 55.6g. Molar mass of Ca ($M_{Ca}$) = 40.08g/mol.
Moles of Ca ($n_{Ca}$) = $\frac{55.6g}{40.08g/mol}\approx1.39mol$

Step2: Calculate moles of S

Mass of S = 44.4g. Molar mass of S ($M_{S}$) = 32.07g/mol.
Moles of S ($n_{S}$) = $\frac{44.4g}{32.07g/mol}\approx1.38mol$

Step3: Find mole - ratio

Divide each mole value by the smaller mole value (1.38mol).
For Ca: $\frac{1.39mol}{1.38mol}\approx1$
For S: $\frac{1.38mol}{1.38mol} = 1$

Answer:

A. $CaS$